How can i evaluate the integral : Pi/2 $(1-3) Sqrt (t^2-1)dt

[Riazy]

Homework Statement

The question scanned

Homework Equations

The Attempt at a Solution

I don't really know what to do

 

[craig16]

Probably try trig sub, with t equal to sec(theta)
don't forget to tack on a sec(theta)tan(theta) when you change dt to dtheta

Edit: I meant sec(theta)tan(theta) not tan^2(theta)

so it becomes

Integral of sec(theta)tan^{2}(theta) dtheta

 

[HallsofIvy]

sin^2(t)+ cos^2(t)=1 so that, dividing by cos^2(t), tan^2(t)+ 1= sec^2(t). That is,
\sqrt{tan^2(t)- 1}= sec(t)

And, of course, if x= tan(t), dx= sec^2(t)dt.

 

[pergradus]

sin^2(t)+ cos^2(t)=1 so that, dividing by cos^2(t), tan^2(t)+ 1= sec^2(t). That is,
\sqrt{tan^2(t)- 1}= sec(t)

And, of course, if x= tan(t), dx= sec^2(t)dt.

This is incorrect. The correct substitution is

t = sec(x) and then the radical becomes \sqrt{sec^2(x)- 1}= tan(x)

and dt/dx = sec(x)tan(x). The integral then becomes:

\int{tan(x)sec(x)tan(x)dx} = \int{sec(x)tan^2(x)dx} = \int{sec(x)-sec^3(x)dx}

edit - I don't know why my latex is all screwed up (I ****ING HATE LATEX) but hopefully he can understand what i wrote

 

[dextercioby]

So

I = \frac{\pi}{2}\int\limits_{1}^{3} \sqrt{x^2 - 1} \, dx.

Well, let x=\cosh t first, compute the new limits and the new integral and then at the end use the double angle formula for \sinh t

\sinh^2 t = \frac{1}{2}\left(\cosh 2t - 1\right)